• "Damian" <damianday hotmail.co.uk> Oct 11 2012
• "Aziz K." <aziz.koeksal gmail.com> Oct 11 2012
• "bearophile" <bearophileHUGS lycos.com> Oct 11 2012
• "Damian" <damianday hotmail.co.uk> Oct 11 2012

"Damian" <damianday hotmail.co.uk> writes:

I come from a pascal background and we could use:
div    integral division operator
/      floating point division operator

So my question is, how does D force floating point division on 
integrals?
At the moment i do this, but i was hoping for an easier way:

int n1 = 10, n2 = 2;
float f = cast(float)(cast(float)(n1 / n2));

"Aziz K." <aziz.koeksal gmail.com> writes:

int n1 = 10, n2 = 2;
float f = (n1+0.0f)/n2;

Casting n1 to float would also work, but I hope the compiler is smart  
enough to optimize away the plus expression.

"bearophile" <bearophileHUGS lycos.com> writes:

Damian:

 I come from a pascal background and we could use:
 div    integral division operator
 /      floating point division operator


Two operators for the two different operations is a design better
than C, that is bug-prone.


 So my question is, how does D force floating point division on 
 integrals?
 At the moment i do this, but i was hoping for an easier way:

 int n1 = 10, n2 = 2;
 float f = cast(float)(cast(float)(n1 / n2));


That's not good, it performs an integer division, followed by two
float casts.

Note: float is useful only if you have many of them, or if you
pass/return pairs of them. A single float is not so useful.

A solution:

int n1 = 10, n2 = 2;
const f = n1 / cast(double)n2;

Using type inference is useful, as it doesn't hide an integer
result if your code is wrong.

Bye,
bearophile

"Damian" <damianday hotmail.co.uk> writes:

On Thursday, 11 October 2012 at 15:21:01 UTC, bearophile wrote:
 Damian:

 I come from a pascal background and we could use:
 div    integral division operator
 /      floating point division operator


 Two operators for the two different operations is a design 
 better
 than C, that is bug-prone.


 So my question is, how does D force floating point division on 
 integrals?
 At the moment i do this, but i was hoping for an easier way:

 int n1 = 10, n2 = 2;
 float f = cast(float)(cast(float)(n1 / n2));


 That's not good, it performs an integer division, followed by 
 two
 float casts.

 Note: float is useful only if you have many of them, or if you
 pass/return pairs of them. A single float is not so useful.

 A solution:

 int n1 = 10, n2 = 2;
 const f = n1 / cast(double)n2;

 Using type inference is useful, as it doesn't hide an integer
 result if your code is wrong.

 Bye,
 bearophile


Ah i see, thankyou for the explanation.